3.200 \(\int e^{-2 i \tan ^{-1}(a+b x)} x^2 \, dx\)

Optimal. Leaf size=59 \[ -\frac {2 i (-a+i)^2 \log (-a-b x+i)}{b^3}+\frac {2 (1+i a) x}{b^2}-\frac {i x^2}{b}-\frac {x^3}{3} \]

[Out]

2*(1+I*a)*x/b^2-I*x^2/b-1/3*x^3-2*I*(I-a)^2*ln(I-a-b*x)/b^3

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Rubi [A]  time = 0.05, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5095, 77} \[ \frac {2 (1+i a) x}{b^2}-\frac {2 i (-a+i)^2 \log (-a-b x+i)}{b^3}-\frac {i x^2}{b}-\frac {x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[x^2/E^((2*I)*ArcTan[a + b*x]),x]

[Out]

(2*(1 + I*a)*x)/b^2 - (I*x^2)/b - x^3/3 - ((2*I)*(I - a)^2*Log[I - a - b*x])/b^3

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 5095

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[((d + e*x)^m*(1 -
I*a*c - I*b*c*x)^((I*n)/2))/(1 + I*a*c + I*b*c*x)^((I*n)/2), x] /; FreeQ[{a, b, c, d, e, m, n}, x]

Rubi steps

\begin {align*} \int e^{-2 i \tan ^{-1}(a+b x)} x^2 \, dx &=\int \frac {x^2 (1-i a-i b x)}{1+i a+i b x} \, dx\\ &=\int \left (\frac {2 i (-i+a)}{b^2}-\frac {2 i x}{b}-x^2-\frac {2 i (-i+a)^2}{b^2 (-i+a+b x)}\right ) \, dx\\ &=\frac {2 (1+i a) x}{b^2}-\frac {i x^2}{b}-\frac {x^3}{3}-\frac {2 i (i-a)^2 \log (i-a-b x)}{b^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 55, normalized size = 0.93 \[ \frac {b x \left (6 i a-b^2 x^2-3 i b x+6\right )-6 i (a-i)^2 \log (-a-b x+i)}{3 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/E^((2*I)*ArcTan[a + b*x]),x]

[Out]

(b*x*(6 + (6*I)*a - (3*I)*b*x - b^2*x^2) - (6*I)*(-I + a)^2*Log[I - a - b*x])/(3*b^3)

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fricas [A]  time = 0.53, size = 53, normalized size = 0.90 \[ -\frac {b^{3} x^{3} + 3 i \, b^{2} x^{2} + 6 \, {\left (-i \, a - 1\right )} b x - {\left (-6 i \, a^{2} - 12 \, a + 6 i\right )} \log \left (\frac {b x + a - i}{b}\right )}{3 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(1+I*(b*x+a))^2*(1+(b*x+a)^2),x, algorithm="fricas")

[Out]

-1/3*(b^3*x^3 + 3*I*b^2*x^2 + 6*(-I*a - 1)*b*x - (-6*I*a^2 - 12*a + 6*I)*log((b*x + a - I)/b))/b^3

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giac [B]  time = 0.14, size = 120, normalized size = 2.03 \[ \frac {2 \, {\left (a^{2} i + 2 \, a - i\right )} \log \left (\frac {1}{\sqrt {{\left (b x + a\right )}^{2} + 1} {\left | b \right |}}\right )}{b^{3}} + \frac {{\left (b i x + a i + 1\right )}^{3} {\left (\frac {3 \, {\left (a b - 2 \, b i\right )} i}{{\left (b i x + a i + 1\right )} b} - \frac {3 \, {\left (a^{2} b^{2} - 6 \, a b^{2} i - 5 \, b^{2}\right )} i^{2}}{{\left (b i x + a i + 1\right )}^{2} b^{2}} - 1\right )}}{3 \, b^{3} i^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(1+I*(b*x+a))^2*(1+(b*x+a)^2),x, algorithm="giac")

[Out]

2*(a^2*i + 2*a - i)*log(1/(sqrt((b*x + a)^2 + 1)*abs(b)))/b^3 + 1/3*(b*i*x + a*i + 1)^3*(3*(a*b - 2*b*i)*i/((b
*i*x + a*i + 1)*b) - 3*(a^2*b^2 - 6*a*b^2*i - 5*b^2)*i^2/((b*i*x + a*i + 1)^2*b^2) - 1)/(b^3*i^3)

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maple [B]  time = 0.06, size = 143, normalized size = 2.42 \[ -\frac {x^{3}}{3}-\frac {i x^{2}}{b}+\frac {2 i a x}{b^{2}}+\frac {2 x}{b^{2}}+\frac {2 \arctan \left (b x +a \right ) a^{2}}{b^{3}}-\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2}}{b^{3}}-\frac {2 \arctan \left (b x +a \right )}{b^{3}}-\frac {4 i \arctan \left (b x +a \right ) a}{b^{3}}+\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b^{3}}-\frac {2 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(1+I*(b*x+a))^2*(1+(b*x+a)^2),x)

[Out]

-1/3*x^3-I*x^2/b+2*I/b^2*a*x+2*x/b^2+2/b^3*arctan(b*x+a)*a^2-I/b^3*ln(b^2*x^2+2*a*b*x+a^2+1)*a^2-2/b^3*arctan(
b*x+a)-4*I/b^3*arctan(b*x+a)*a+I/b^3*ln(b^2*x^2+2*a*b*x+a^2+1)-2/b^3*ln(b^2*x^2+2*a*b*x+a^2+1)*a

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maxima [A]  time = 0.32, size = 52, normalized size = 0.88 \[ -\frac {b^{2} x^{3} + 3 i \, b x^{2} + 6 \, {\left (-i \, a - 1\right )} x}{3 \, b^{2}} + \frac {{\left (-2 i \, a^{2} - 4 \, a + 2 i\right )} \log \left (i \, b x + i \, a + 1\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(1+I*(b*x+a))^2*(1+(b*x+a)^2),x, algorithm="maxima")

[Out]

-1/3*(b^2*x^3 + 3*I*b*x^2 + 6*(-I*a - 1)*x)/b^2 + (-2*I*a^2 - 4*a + 2*I)*log(I*b*x + I*a + 1)/b^3

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mupad [B]  time = 0.54, size = 90, normalized size = 1.53 \[ -\ln \left (x+\frac {a-\mathrm {i}}{b}\right )\,\left (\frac {4\,a}{b^3}+\frac {\left (2\,a^2-2\right )\,1{}\mathrm {i}}{b^3}\right )+x^2\,\left (\frac {a-\mathrm {i}}{2\,b}-\frac {a+1{}\mathrm {i}}{2\,b}\right )-\frac {x^3}{3}-\frac {x\,\left (\frac {a-\mathrm {i}}{b}-\frac {a+1{}\mathrm {i}}{b}\right )\,\left (a-\mathrm {i}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*((a + b*x)^2 + 1))/(a*1i + b*x*1i + 1)^2,x)

[Out]

x^2*((a - 1i)/(2*b) - (a + 1i)/(2*b)) - log(x + (a - 1i)/b)*((4*a)/b^3 + ((2*a^2 - 2)*1i)/b^3) - x^3/3 - (x*((
a - 1i)/b - (a + 1i)/b)*(a - 1i))/b

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sympy [A]  time = 0.36, size = 53, normalized size = 0.90 \[ - \frac {x^{3}}{3} - x \left (- \frac {2 i a}{b^{2}} - \frac {2}{b^{2}}\right ) - \frac {i x^{2}}{b} - \frac {2 i \left (a - i\right )^{2} \log {\left (i a + i b x + 1 \right )}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(1+I*(b*x+a))**2*(1+(b*x+a)**2),x)

[Out]

-x**3/3 - x*(-2*I*a/b**2 - 2/b**2) - I*x**2/b - 2*I*(a - I)**2*log(I*a + I*b*x + 1)/b**3

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